Examples of solving equations with large powers. Equations of higher degrees

Let's consider solving equations with one variable of degree higher than the second.

The degree of the equation P(x) = 0 is the degree of the polynomial P(x), i.e. the greatest of the powers of its terms with a coefficient not equal to zero.

So, for example, the equation (x 3 – 1) 2 + x 5 = x 6 – 2 has the fifth degree, because after the operations of opening the brackets and bringing similar ones, we obtain the equivalent equation x 5 – 2x 3 + 3 = 0 of the fifth degree.

Let us recall the rules that will be needed to solve equations of degree higher than two.

Statements about the roots of a polynomial and its divisors:

1. Polynomial nth degrees has a number of roots not exceeding n, and roots of multiplicity m occur exactly m times.

2. A polynomial of odd degree has at least one real root.

3. If α is the root of P(x), then P n (x) = (x – α) · Q n – 1 (x), where Q n – 1 (x) is a polynomial of degree (n – 1).

5. The reduced polynomial with integer coefficients cannot have fractional rational roots.

6. For a third degree polynomial

P 3 (x) = ax 3 + bx 2 + cx + d one of two things is possible: either it is decomposed into the product of three binomials

Р 3 (x) = а(х – α)(х – β)(х – γ), or decomposes into the product of a binomial and a square trinomial Р 3 (x) = а(х – α)(х 2 + βх + γ ).

7. Any polynomial of the fourth degree can be expanded into the product of two square trinomials.

8. A polynomial f(x) is divisible by a polynomial g(x) without a remainder if there is a polynomial q(x) such that f(x) = g(x) · q(x). To divide polynomials, the “corner division” rule is used.

9. For the polynomial P(x) to be divisible by a binomial (x – c), it is necessary and sufficient that the number c be the root of P(x) (Corollary of Bezout’s theorem).

10. Vieta's theorem: If x 1, x 2, ..., x n are real roots of the polynomial

P(x) = a 0 x n + a 1 x n - 1 + ... + a n, then the following equalities hold:

x 1 + x 2 + … + x n = -a 1 /a 0,

x 1 x 2 + x 1 x 3 + ... + x n – 1 x n = a 2 /a 0,

x 1 x 2 x 3 + … + x n – 2 x n – 1 x n = -a 3 / a 0,

x 1 · x 2 · x 3 · x n = (-1) n a n / a 0 .

Solving Examples

Example 1.

Find the remainder of division P(x) = x 3 + 2/3 x 2 – 1/9 by (x – 1/3).

Solution.

By corollary to Bezout’s theorem: “The remainder of a polynomial divided by a binomial (x – c) is equal to the value of the polynomial of c.” Let's find P(1/3) = 0. Therefore, the remainder is 0 and the number 1/3 is the root of the polynomial.

Answer: R = 0.

Example 2.

Divide with a “corner” 2x 3 + 3x 2 – 2x + 3 by (x + 2). Find the remainder and incomplete quotient.

Solution:

2x 3 + 3x 2 – 2x + 3| x + 2

2x 3 + 4 x 2 2x 2 – x

X 2 – 2 x

Answer: R = 3; quotient: 2x 2 – x.

Basic methods for solving equations higher degrees

1. Introduction of a new variable

The method of introducing a new variable is already familiar from the example of bi quadratic equations. It consists in the fact that to solve the equation f(x) = 0, a new variable (substitution) t = x n or t = g(x) is introduced and f(x) is expressed through t, obtaining a new equation r(t). Then solving the equation r(t), the roots are found:

(t 1, t 2, …, t n). After this, a set of n equations q(x) = t 1 , q(x) = t 2 , … , q(x) = t n is obtained, from which the roots of the original equation are found.

Example 1.

(x 2 + x + 1) 2 – 3x 2 – 3x – 1 = 0.

Solution:

(x 2 + x + 1) 2 – 3(x 2 + x) – 1 = 0.

(x 2 + x + 1) 2 – 3(x 2 + x + 1) + 3 – 1 = 0.

Substitution (x 2 + x + 1) = t.

t 2 – 3t + 2 = 0.

t 1 = 2, t 2 = 1. Reverse substitution:

x 2 + x + 1 = 2 or x 2 + x + 1 = 1;

x 2 + x - 1 = 0 or x 2 + x = 0;

Answer: From the first equation: x 1, 2 = (-1 ± √5)/2, from the second: 0 and -1.

2. Factorization by grouping and abbreviated multiplication formulas

The basis of this method is also not new and consists in grouping terms in such a way that each group contains a common factor. To do this, sometimes it is necessary to use some artificial techniques.

Example 1.

x 4 – 3x 2 + 4x – 3 = 0.

Solution.

Let's imagine - 3x 2 = -2x 2 – x 2 and group:

(x 4 – 2x 2) – (x 2 – 4x + 3) = 0.

(x 4 – 2x 2 +1 – 1) – (x 2 – 4x + 3 + 1 – 1) = 0.

(x 2 – 1) 2 – 1 – (x – 2) 2 + 1 = 0.

(x 2 – 1) 2 – (x – 2) 2 = 0.

(x 2 – 1 – x + 2)(x 2 – 1 + x - 2) = 0.

(x 2 – x + 1)(x 2 + x – 3) = 0.

x 2 – x + 1 = 0 or x 2 + x – 3 = 0.

Answer: There are no roots in the first equation, from the second: x 1, 2 = (-1 ± √13)/2.

3. Factorization method uncertain coefficients

The essence of the method is that the original polynomial is factorized with unknown coefficients. Using the property that polynomials are equal if their coefficients are equal for the same powers, the unknown expansion coefficients are found.

Example 1.

x 3 + 4x 2 + 5x + 2 = 0.

Solution.

A polynomial of degree 3 can be expanded into the product of linear and quadratic factors.

x 3 + 4x 2 + 5x + 2 = (x – a)(x 2 + bx + c),

x 3 + 4x 2 + 5x + 2 = x 3 +bx 2 + cx – ax 2 – abx – ac,

x 3 + 4x 2 + 5x + 2 = x 3 + (b – a)x 2 + (cx – ab)x – ac.

Having solved the system:

(b – a = 4,
(c – ab = 5,
(-ac = 2,

(a = -1,
(b = 3,
(c = 2, i.e.

x 3 + 4x 2 + 5x + 2 = (x + 1)(x 2 + 3x + 2).

The roots of the equation (x + 1)(x 2 + 3x + 2) = 0 are easy to find.

Answer: -1; -2.

4. Method of selecting a root using the highest and free coefficient

The method is based on the application of theorems:

1) Every integer root of a polynomial with integer coefficients is a divisor of the free term.

2) In order for the irreducible fraction p/q (p is an integer, q is a natural number) to be the root of an equation with integer coefficients, it is necessary that the number p be an integer divisor of the free term a 0, and q be a natural divisor of the leading coefficient.

Example 1.

6x 3 + 7x 2 – 9x + 2 = 0.

Solution:

6: q = 1, 2, 3, 6.

Therefore, p/q = ±1, ±2, ±1/2, ±1/3, ±2/3, ±1/6.

Having found one root, for example – 2, we will find other roots using corner division, the method of indefinite coefficients or Horner’s scheme.

Answer: -2; 1/2; 1/3.

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“Methods for solving equations of higher degrees”

( Kiselev Readings)

Mathematics teacher Afanasyeva L.A.

MKOU Verkhnekarachskaya secondary school

Gribanovsky district, Voronezh region

2015

Mathematics education received in secondary school, is essential component general education And general culture modern man.

The famous German mathematician Courant wrote: “For two years more than a thousand years possession of some, not too superficial, knowledge in the field of mathematics was necessary integral part into everyone's intellectual inventory educated person" And among this knowledge there is no last place belongs to the ability to solve equations.

Already in ancient times, people realized how important it was to learn to solve algebraic equations. About 4000 years ago, Babylonian scientists knew how to solve a quadratic equation and solved systems of two equations, one of which was of the second degree. With the help of equations, various problems of land surveying, architecture and military affairs were solved; many and varied questions of practice and natural science were reduced to them, since the precise language of mathematics allows one to simply express facts and relationships that, when stated in ordinary language, may seem confusing and complex. Equation one of the most important concepts mathematics. The development of methods for solving equations, starting from the birth of mathematics as a science, for a long time was the main subject of algebra. And today in mathematics lessons, starting from the first stage of education, solving equations various types much attention is paid.

Universal formula for finding roots algebraic equation There is no nth degree. Many, of course, had the tempting idea of finding for any degree n formulas that would express the roots of the equation through its coefficients, that is, would solve the equation in radicals. However, the “dark Middle Ages” turned out to be as gloomy as possible in relation to the problem under discussion - for seven whole centuries no one found the required formulas! Only in the 16th century did Italian mathematicians manage to advance further - to find formulas for n =3 And n =4 . At the same time, the question about general decision equations of the 3rd degree were studied by Scipio Dal Ferro, his student Fiori and Tartaglia. In 1545, the book of the Italian mathematician D Cardano “The Great Art, or On the Rules of Algebra” was published, where, along with other questions of algebra, general methods solving cubic equations, as well as a method for solving equations of the 4th degree, discovered by his student L. Ferrari. A complete presentation of issues related to the solution of equations of 3rd to 4th degrees was given by F. Viet. And in the 20s of the 19th century, the Norwegian mathematician N. Abel proved that the roots of equations of the 5th and higher degrees cannot be expressed in terms of radicals.

The process of finding solutions to an equation usually involves replacing the equation with an equivalent one. Replacing the equation with an equivalent one is based on the use of four axioms:

1. If equal values increase by the same number, the results will be equal.

2. If you subtract the same number from equal quantities, the results will be equal.

3. If equal values are multiplied by the same number, the results will be equal.

4. If equal quantities are divided by the same number, the results will be equal.

Since the left side of the equation P(x) = 0 is a polynomial of nth degree, it is useful to recall the following statements:

Statements about the roots of a polynomial and its divisors:

1. A polynomial of nth degree has a number of roots not exceeding n, and roots of multiplicity m occur exactly m times.

2. A polynomial of odd degree has at least one real root.

3. If α is the root of P(x), then P n (x) = (x - α) Q n - 1 (x), where Q n - 1 (x) is a polynomial of degree (n - 1).

4. Every integer root of a polynomial with integer coefficients is a divisor of the free term.

5. The reduced polynomial with integer coefficients cannot have fractional rational roots.

6. For a third degree polynomial

P 3 (x) = ax 3 + bx 2 + cx + d one of two things is possible: either it is decomposed into the product of three binomials

P 3 (x) = a (x - α)(x - β)(x - γ), or decomposes into the product of a binomial and a square trinomial P 3 (x) = a(x - α)(x 2 + βx + γ ).

7. Any polynomial of the fourth degree can be expanded into the product of two square trinomials.

8. A polynomial f (x) is divisible by a polynomial g(x) without a remainder if there is a polynomial q(x) such that f(x) = g(x) q(x). To divide polynomials, the “corner division” rule is used.

9. For the polynomial P(x) to be divisible by a binomial (x – c), it is necessary and sufficient that c be the root of P(x) (Corollary of Bezout’s theorem).

10. Vieta's theorem: If x 1, x 2, ..., x n are real roots of the polynomial

P(x) = a 0 x n + a 1 x n - 1 + ... + a n, then the following equalities hold:

x 1 + x 2 + … + x n = -a 1 /a 0,

x 1 x 2 + x 1 x 3 + ... + x n - 1 x n = a 2 /a 0,

x 1 x 2 x 3 + ... + x n - 2 x n - 1 x n = -a 3 /a 0,

x 1 x 2 x 3 x n = (-1) n a n /a 0 .

Solving Examples

Example 1 . Find the remainder of division P(x) = x 3 + 2/3 x 2 – 1/9 by (x – 1/3).

Solution. According to the corollary of Bezout’s theorem: “The remainder of a polynomial divided by a binomial (x - c) is equal to the value of the polynomial of c.” Let's find P(1/3) = 0. Therefore, the remainder is 0 and the number 1/3 is the root of the polynomial.

Answer: R = 0.

Example 2 . Divide with a “corner” 2x 3 + 3x 2 – 2x + 3 by (x + 2). Find the remainder and incomplete quotient.

Solution:

2x 3 + 3x 2 – 2x + 3| x + 2

2x 3 + 4x 2 2x 2 – x

X 2 – 2x

Answer: R = 3; quotient: 2x 2 – x.

Basic methods for solving higher degree equations

1. Introduction of a new variable

The method of introducing a new variable is that to solve the equation f(x) = 0, a new variable (substitution) t = x n or t = g(x) is introduced and f(x) is expressed through t, obtaining a new equation r(t) . Then solving the equation r(t), the roots are found: (t 1, t 2, ..., t n). After this, a set of n equations q(x) = t 1 , q(x) = t 2 , … , q(x) = t n is obtained, from which the roots of the original equation are found.

Example;(x 2 + x + 1) 2 – 3x 2 – 3x – 1 = 0.

Solution: (x 2 + x + 1) 2 – 3x 2 – 3x – 1 = 0.

(x 2 + x + 1) 2 – 3(x 2 + x + 1) + 3 – 1 = 0.

Substitution (x 2 + x + 1) = t.

t 2 – 3t + 2 = 0.

t 1 = 2, t 2 = 1. Reverse substitution:

x 2 + x + 1 = 2 or x 2 + x + 1 = 1;

x 2 + x - 1 = 0 or x 2 + x = 0;

From the first equation: x 1, 2 = (-1 ± √5)/2, from the second: 0 and -1.

The method of introducing a new variable is used in solving returnable equations, that is, equations of the form a 0 x n + a 1 x n – 1 + .. + a n – 1 x + a n =0, in which the coefficients of the terms of the equation, equally spaced from the beginning and end, are equal.

2. Factorization by grouping and abbreviated multiplication formulas

The basis of this method is to group the terms so that each group contains a common factor. To do this, sometimes it is necessary to use some artificial techniques.

Example: x 4 – 3x 2 + 4x – 3 = 0.

Solution. Imagine - 3x 2 = -2x 2 – x 2 and group:

(x 4 - 2x 2) – (x 2 - 4x + 3) = 0.

(x 4 – 2x 2 +1 – 1) – (x 2 – 4x + 3 + 1 – 1) = 0.

(x 2 – 1) 2 – 1 – (x – 2) 2 + 1 = 0.

(x 2 – 1) 2 – (x – 2) 2 = 0.

(x 2 – 1 – x + 2)(x 2 – 1 + x – 2) = 0.

(x 2 – x + 1)(x 2 + x – 3) = 0.

x 2 – x + 1 = 0 or x 2 + x – 3 = 0.

There are no roots in the first equation, from the second: x 1, 2 = (-1 ± √13)/2.

3. Factorization by the method of undetermined coefficients

Example: x 3 + 4x 2 + 5x + 2 = 0.

Solution. A polynomial of degree 3 can be expanded into the product of linear and quadratic factors.

x 3 + 4x 2 + 5x + 2 = (x - a)(x 2 + bx + c),

x 3 + 4x 2 + 5x + 2 = x 3 +bx 2 + cx - ax 2 - abx - ac,

x 3 + 4x 2 + 5x + 2 = x 3 + (b – a)x 2 + (c – ab)x – ac.

Having solved the system:

we get

x 3 + 4x 2 + 5x + 2 = (x + 1)(x 2 + 3x + 2).

The roots of the equation (x + 1)(x 2 + 3x + 2) = 0 are easy to find.

Answer: -1; -2.

4. Method of selecting a root using the highest and free coefficient

The method is based on the application of theorems:

1) Every integer root of a polynomial with integer coefficients is a divisor of the free term.

2) In order for the irreducible fraction p/q (p - integer, q - natural) to be the root of an equation with integer coefficients, it is necessary that the number p be an integer divisor of the free term a 0, and q - a natural divisor of the leading coefficient.

Example: 6x 3 + 7x 2 - 9x + 2 = 0.

Solution:

2: p = ±1, ±2

6: q = 1, 2, 3, 6.

Therefore, p/q = ±1, ±2, ±1/2, ±1/3, ±2/3, ±1/6.

Having found one root, for example – 2, we will find other roots using corner division, the method of indefinite coefficients or Horner’s scheme.

Answer: -2; 1/2; 1/3.

5. Graphic method.

This method consists of constructing graphs and using the properties of functions.

Example: x 5 + x – 2 = 0

Let's imagine the equation in the form x 5 = - x + 2. The function y = x 5 is increasing, and the function y = - x + 2 is decreasing. This means that the equation x 5 + x – 2 = 0 has a single root -1.

6.Multiplying an equation by a function.

Sometimes solving an algebraic equation is significantly easier if you multiply both sides by a certain function - a polynomial in the unknown. At the same time, we must remember that it is possible that extra roots may appear—the roots of the polynomial by which the equation was multiplied. Therefore, you must either multiply by a polynomial that has no roots and get an equivalent equation, or multiply by a polynomial that has roots, and then each of these roots must be substituted into the original equation and determine whether this number is its root.

Example. Solve the equation:

X 8 – X 6 + X 4 – X 2 + 1 = 0. (1)

Solution: Multiplying both sides of the equation by the polynomial X 2 + 1, which has no roots, we obtain the equation:

(X 2 +1) (X 8 – X 6 + X 4 – X 2 + 1) = 0 (2)
equivalent to equation (1). Equation (2) can be written as:

X 10 + 1= 0 (3)
It is clear that equation (3) does not have real roots, so equation (1) does not have them.

Answer: no solutions.

In addition to the above methods for solving equations of higher degrees, there are others. For example, highlighting a complete square, Horner's scheme, representing a fraction as two fractions. From common methods solutions to equations of higher degrees, which occur most often, use: the method of factoring the left side of the equation;

variable replacement method (method of introducing a new variable); graphic method. We introduce these methods to 9th grade students when studying the topic “The Whole Equation and Its Roots.” In the textbook Algebra 9 (authors Makarychev Yu.N., Mindyuk N.G., etc.) of recent years of publication, the main methods for solving equations of higher degrees are discussed in sufficient detail. In addition, in the section “For those who want to know more,” in my opinion, material on the application of theorems on the root of a polynomial and entire roots of an entire equation when solving equations of higher degrees is presented in an accessible manner. Well-prepared students study this material with interest and then present the solved equations to their classmates.

Almost everything that surrounds us is connected to one degree or another with mathematics. And achievements in physics, technology, information technology only confirm this. And what is very important is that solving many practical problems comes down to solving various types of equations that you need to learn how to solve.

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Introduction

Solving algebraic equations of higher degrees with one unknown is one of the most difficult and ancient mathematical problems. The most outstanding mathematicians of antiquity dealt with these problems.

Solving equations of the nth degree is an important task for modern mathematics. There is quite a lot of interest in them, since these equations are closely related to the search for the roots of equations that are not covered in the school mathematics curriculum.

Problem: Students’ lack of skills in solving equations of higher degrees in various ways prevents them from successfully preparing for final certification in mathematics and mathematical Olympiads, and training in a specialized mathematics class.

The listed facts determined relevance our work “Solving equations of higher degrees”.

Knowledge of the simplest methods of solving equations of the nth degree reduces the time for completing a task, on which the result of the work and the quality of the learning process depend.

Goal of the work: studying known methods solving equations of higher degrees and identifying the most accessible of them for practical application.

Based on the goal, the work identifies the following: tasks:

Study literature and Internet resources on this topic;

Get acquainted with historical facts related to this topic;

Describe different ways to solve higher degree equations

compare the degree of complexity of each of them;

Introduce classmates to ways of solving equations of higher degrees;

Create a selection of equations for the practical application of each of the considered methods.

Object of study- equations of higher degrees with one variable.

Subject of study- methods for solving equations of higher degrees.

Hypothesis: There is no general method or single algorithm that allows one to find solutions to equations of the nth degree in a finite number of steps.

Research methods:

- bibliographic method (analysis of literature on the research topic);

- classification method;

- method of qualitative analysis.

Theoretical significance research consists of systematizing methods for solving equations of higher degrees and describing their algorithms.

Practical significance- presented material on this topic and development teaching aid for students on this topic.

1. EQUATIONS OF HIGHER DEGREES

1.1 Concept of nth degree equation

Definition 1. An equation of the nth degree is an equation of the form

a 0 xⁿ+a 1 x n -1 +a 2 xⁿ - ²+…+a n -1 x+a n = 0, where coefficients a 0, a 1, a 2…, a n -1, a n- any real numbers, and ,a 0 ≠ 0 .

Polynomial a 0 xⁿ+a 1 x n -1 +a 2 xⁿ - ²+…+a n -1 x+a n is called a polynomial of nth degree. The coefficients are distinguished by names: a 0 - senior coefficient; a n is a free member.

Definition 2. Solutions or roots for given equation are all the values of the variable X, which turn this equation into a true numerical equality or, for which the polynomial a 0 xⁿ+a 1 x n -1 +a 2 xⁿ - ²+…+a n -1 x+a n goes to zero. This variable value X also called the root of a polynomial. Solving an equation means finding all its roots or establishing that there are none.

If a 0 = 1, then such an equation is called a reduced integer rational equation n th degrees.

For equations of the third and fourth degree, there are Cardano and Ferrari formulas that express the roots of these equations through radicals. It turned out that in practice they are rarely used. Thus, if n ≥ 3, and the coefficients of the polynomial are arbitrary real numbers, then finding the roots of the equation is not an easy task. However, in many special cases this problem is completely solved. Let's look at some of them.

1.2 Historical facts solving higher degree equations

A universal formula for finding the roots of an algebraic equation nth no degree. Many, of course, had the tempting idea of finding, for any degree n, formulas that would express the roots of the equation through its coefficients, that is, solve the equation in radicals.

Only in the 16th century did Italian mathematicians manage to advance further - to find formulas for n = 3 and n = 4. At the same time, Scipio, Dahl, Ferro and his students Fiori and Tartaglia were studying the question of the general solution of equations of the 3rd degree.

In 1545, the book of the Italian mathematician D. Cardano “Great Art, or on the Rules of Algebra” was published, where, along with other questions of algebra, general methods for solving cubic equations are considered, as well as a method for solving equations of the 4th degree, discovered by his student L. Ferrari.

A complete presentation of issues related to the solution of equations of the 3rd and 4th degrees was given by F. Viet.

In the 20s of the 19th century, the Norwegian mathematician N. Abel proved that the roots of equations of the fifth degree cannot be expressed in terms of radicals.

The study revealed that modern science There are many ways to solve equations of the nth degree.

The result of the search for methods for solving equations of higher degrees that cannot be solved by the methods considered in school curriculum, methods based on the application of Vieta’s theorem (for equations of degree n>2), Bezout's theorems, Horner's schemes, as well as the Cardano and Ferrari formula for solving cubic and quartic equations.

The work presents methods for solving equations and their types, which became a discovery for us. These include the method of indefinite coefficients, the selection of the full degree, symmetric equations.

2. SOLUTION OF ENTIRE EQUATIONS OF HIGHER DEGREES WITH INTEGER COEFFICIENTS

2.1 Solving 3rd degree equations. Formula D. Cardano

Consider equations of the form x 3 +px+q=0. Let's transform the equation general view to the form: x 3 +px 2 +qx+r=0. Let's write down the formula for the cube of the sum; Let's add it to the original equality and replace it with y. We get the equation: y 3 + (q -) (y -) + (r - =0. After transformations, we have: y 2 +py + q=0. Now, let’s write down the sum cube formula again:

(a+b) 3 =a 3 + 3a 2 b + 3ab 2 +b 3 = a 3 +b 3 + 3ab (a + b), replace ( a+b)on x, we get the equation x 3 - 3abx - (a 3 +b 3) = 0. Now we can see that the original equation is equivalent to the system: and Solving the system, we get:

We have obtained a formula for solving the above 3rd degree equation. It bears the name of the Italian mathematician Cardano.

Let's look at an example. Solve the equation: .

We have R= 15 and q= 124, then using the Cardano formula we calculate the root of the equation

Conclusion: this formula is good, but not suitable for solving all cubic equations. At the same time, it is cumbersome. Therefore, in practice it is rarely used.

But anyone who masters this formula can use it when solving third-degree equations on the Unified State Exam.

2.2 Vieta's theorem

From a mathematics course we know this theorem for a quadratic equation, but few people know that it is also used to solve higher-order equations.

Consider the equation:

Let's factorize the left side of the equation and divide by ≠ 0.

Let's transform the right side of the equation to the form

; It follows that we can write the following equalities into the system:

The formulas derived by Viète for quadratic equations and demonstrated by us for equations of the 3rd degree are also true for polynomials of higher degrees.

Let's solve the cubic equation:

Conclusion: this method is universal and easy enough for students to understand, since Vieta’s theorem is familiar to them from the school curriculum for n = 2. At the same time, in order to find the roots of equations using this theorem, you must have good computational skills.

2.3 Bezout's theorem

This theorem is named after the 18th century French mathematician J. Bezout.

Theorem. If the equation a 0 xⁿ+a 1 x n -1 +a 2 xⁿ - ²+…+a n -1 x+a n = 0, in which all coefficients are integers, and the free term is non-zero and has an integer root, then this root is a divisor of the free term.

Considering that on the left side of the equation there is a polynomial of nth degree, the theorem has another interpretation.

Theorem. When dividing a polynomial of nth degree with respect to x by binomial x-a the remainder is equal to the value of the dividend when x = a. (letter a can denote any real or imaginary number, i.e. any complex number).

Proof: let f(x) denotes an arbitrary polynomial of the nth degree with respect to the variable x and let, when divided by a binomial ( x-a) turned out in private q(x), and the remainder R. It's obvious that q(x) there will be some polynomial (n - 1)th degree relative x, and the remainder R will be a constant value, i.e. independent of x.

If the remainder R was a polynomial of the first degree with respect to x, then this would mean that the division failed. So, R from x does not depend. By definition of division we obtain the identity: f(x)=(x-a) q(x)+R.

The equality is true for any value of x, which means it is also true for x=a, we get: f(a)=(a-a) q(a)+R. Symbol f(a) denotes the value of the polynomial f (x) at x=a, q(a) stands for value q(x) at x=a. Remainder R remained the same as it was before, because R from x does not depend. Work ( x-a) q(a) = 0, since the factor ( x-a) = 0, and the multiplier q(a) There is certain number. Therefore, from the equality we get: f(a)= R, etc.

Example 1. Find the remainder of a polynomial x 3 - 3x 2 + 6x- 5 per binomial

x- 2. By Bezout’s theorem : R=f(2) = 23-322 + 62 -5=3. Answer: R= 3.

Note that Bezout’s theorem is important not so much in itself as for its consequences. (Annex 1)

Let us dwell on the consideration of some techniques for applying Bezout’s theorem to solving practical problems. It should be noted that when solving equations using Bezout’s theorem, it is necessary:

Find all integer divisors of the free term;

Find at least one root of the equation from these divisors;

Divide the left side of the equation by (Ha);

Write down the product of the divisor and the quotient on the left side of the equation;

Solve the resulting equation.

Let's look at the example of solving the equation x 3 + 4X 2 + x - 6 = 0 .

Solution: find the divisors of the free term ±1 ; ± 2; ± 3; ± 6. Let's calculate the values at x= 1, 1 3 + 41 2 + 1- 6=0. Divide the left side of the equation by ( X- 1). Let’s do the division using a “corner” and get:

Conclusion: Bezout’s theorem is one of the methods that we consider in our work, studied in the elective curriculum. It is difficult to understand, because in order to master it, you need to know all the consequences from it, but at the same time, Bezout’s theorem is one of the main assistants for students on the Unified State Exam.

2.4 Horner scheme

To divide a polynomial by a binomial x-α you can use a special simple technique invented by English mathematicians of the 17th century, later called Horner’s scheme. In addition to finding the roots of equations, using Horner's scheme you can more simply calculate their values. To do this, you need to substitute the value of the variable into the polynomial Pn (x)=a 0 xn+a 1 x n-1 +a 2 xⁿ - ²+…++a n -1 x+a n. (1)

Consider dividing the polynomial (1) by the binomial x-α.

Let us express the coefficients of the incomplete quotient b 0 xⁿ - ¹+ b 1 xⁿ - ²+ b 2 xⁿ - ³+…+ bn -1 and the remainder r through the coefficients of the polynomial Pn( x) and number α. b 0 =a 0 , b 1 = α b 0 +a 1 , b 2 = α b 1 +a 2 …, bn -1 =

= α bn -2 +a n -1 = α bn -1 +a n .

Calculations according to Horner’s scheme are presented in the following table:

	A 0	a 1	a 2 ,
	b 0 =a 0	b 1 = α b 0 +a 1	b 2 = α b 1 +a 2		r=α b n-1 +a n

Because the r=Pn(α), then α is the root of the equation. In order to check whether α is a multiple root, Horner’s scheme can be applied to the quotient b 0 x+ b 1 x+…+ bn -1 according to the table. If in the column under bn -1 the result is 0 again, which means α is a multiple root.

Let's look at an example: solve the equation X 3 + 4X 2 + x - 6 = 0.

Let us apply to the left side of the equation the factorization of the polynomial on the left side of the equation, Horner's scheme.

Solution: find the divisors of the free term ± 1; ± 2; ± 3; ± 6.


				6 ∙ 1 + (-6) = 0

The coefficients of the quotient are the numbers 1, 5, 6, and the remainder r = 0.

Means, X 3 + 4X 2 + X - 6 = (X - 1) (X 2 + 5X + 6) = 0.

From here: X- 1 = 0 or X 2 + 5X + 6 = 0.

X = 1, X 1 = -2; X 2 = -3. Answer: 1,- 2, - 3.

Conclusion: thus, on one equation we have shown the use of two in various ways factorization of polynomials. In our opinion, Horner's scheme is the most practical and economical.

2.5 Solving 4th degree equations. Ferrari method

Cardano's student Ludovic Ferrari discovered a way to solve a fourth-degree equation. The Ferrari method consists of two stages.

Stage I: equations of the form are represented as the product of two square trinomials; this follows from the fact that the equation is of the 3rd degree and has at least one solution.

Stage II: the resulting equations are solved using factorization, but in order to find the required factorization, cubic equations have to be solved.

The idea is to represent the equations in the form A 2 =B 2, where A= x 2 +s,

B-linear function of x. Then it remains to solve the equations A = ±B.

For clarity, consider the equation: Isolating the 4th degree, we get: For any d the expression will be a perfect square. Add to both sides of the equation we get

On the left side there is a complete square, you can pick up d, so that the right side of (2) also becomes a complete square. Let's imagine that we have achieved this. Then our equation looks like this:

Finding the root will not be difficult later. To choose the right d it is necessary that the discriminant of the right side of (3) becomes zero, i.e.

So to find d, we need to solve this 3rd degree equation. This auxiliary equation is called resolvent.

We easily find the whole root of the resolvent: d = 1

Substituting the equation into (1) we get

Conclusion: the Ferrari method is universal, but complex and cumbersome. At the same time, if the solution algorithm is clear, then 4th degree equations can be solved using this method.

2.6 Method of undetermined coefficients

The success of solving an equation of the 4th degree using the Ferrari method depends on whether we solve the resolvent - an equation of the 3rd degree, which, as we know, is not always possible.

The essence of the method of indefinite coefficients is that the type of factors into which a given polynomial is decomposed is guessed, and the coefficients of these factors (also polynomials) are determined by multiplying the factors and equating the coefficients at the same powers of the variable.

Example: solve the equation:

Suppose that the left side of our equation can be decomposed into two square trinomials with integer coefficients such that the identical equality is true

Obviously, the coefficients in front of them must be equal to 1, and the free terms must be equal to one + 1, the other - 1.

The coefficients facing the X. Let us denote them by A and and to determine them, we multiply both trinomials on the right side of the equation.

As a result we get:

Equating coefficients at the same degrees X on the left and right sides of equality (1), we obtain a system for finding and

Having solved this system, we will have

So our equation is equivalent to the equation

Having solved it, we get the following roots: .

The method of indefinite coefficients is based on the following statements: any polynomial of the fourth degree in the equation can be decomposed into the product of two polynomials of the second degree; two polynomials are identically equal if and only if their coefficients are equal for the same powers X.

2.7 Symmetric equations

Definition. An equation of the form is called symmetric if the first coefficients on the left of the equation are equal to the first coefficients on the right.

We see that the first coefficients on the left are equal to the first coefficients on the right.

If such an equation has an odd degree, then it has a root X= - 1. Next we can lower the degree of the equation by dividing it by ( x+ 1). It turns out that when dividing a symmetric equation by ( x+ 1) a symmetric equation of even degree is obtained. Proof of the symmetry of the coefficients is presented below. (Appendix 6) Our task is to learn how to solve symmetric equations of even degree.

For example: (1)

Let's solve equation (1), divide by X 2 (to medium degree) = 0.

Let us group terms with symmetric

) + 3(x+ . Let's denote at= x+ , let’s square both sides, hence = at 2 So, 2( at 2 or 2 at 2 + 3 solving the equation, we get at = , at= 3. Next, let's return to replacement x+ = and x+ = 3. We obtain the equations and The first has no solution, and the second has two roots. Answer:.

Conclusion: this type of equation is not often encountered, but if you come across it, then it can be solved easily and simply without resorting to cumbersome calculations.

2.8 Isolation of a full degree

Consider the equation.

The left side is the cube of the sum (x+1), i.e.

We extract the third root from both parts: , then we get

Where is the only root?

RESEARCH RESULTS

Based on the results of the work, we came to the following conclusions:

Thanks to the studied theory, we became acquainted with various methods solving entire equations of higher degrees;

D. Cardano's formula is difficult to use and gives a high probability of making errors in the calculation;

− L. Ferrari’s method allows one to reduce the solution to a fourth-degree equation to a cubic one;

− Bezout’s theorem can be used both for cubic equations and for equations of the fourth degree; it is more understandable and visual when applied to solving equations;

Horner's scheme helps to significantly reduce and simplify calculations in solving equations. In addition to finding the roots, using Horner's scheme you can more simply calculate the values of the polynomials on the left side of the equation;

Of particular interest were the solutions of equations by the method of indefinite coefficients and the solution of symmetric equations.

During research work It was found that students become familiar with the simplest methods of solving equations of the highest degree in elective mathematics classes, starting in the 9th or 10th grade, as well as in special courses at visiting mathematics schools. This fact established as a result of a survey of mathematics teachers of MBOU “Secondary School No. 9” and students showing increased interest in the subject “mathematics”.

The most popular methods for solving equations of higher degrees, which are encountered when solving olympiads, competitive problems and as a result of students preparing for exams, are methods based on the application of Bezout’s theorem, Horner’s scheme and the introduction of a new variable.

Demonstration of the results of research work, i.e. methods for solving equations not taught in the school mathematics curriculum interested my classmates.

Conclusion

Having studied the educational and scientific literature, Internet resources in youth educational forums

Let's consider solving equations with one variable of degree higher than the second.

The degree of the equation P(x) = 0 is the degree of the polynomial P(x), i.e. the greatest of the powers of its terms with a coefficient not equal to zero.

Let us recall the rules that will be needed to solve equations of degree higher than two.

Statements about the roots of a polynomial and its divisors:

1. A polynomial of nth degree has a number of roots not exceeding n, and roots of multiplicity m occur exactly m times.

2. A polynomial of odd degree has at least one real root.

3. If α is the root of P(x), then P n (x) = (x – α) · Q n – 1 (x), where Q n – 1 (x) is a polynomial of degree (n – 1).

5. The reduced polynomial with integer coefficients cannot have fractional rational roots.

6. For a third degree polynomial

P 3 (x) = ax 3 + bx 2 + cx + d one of two things is possible: either it is decomposed into the product of three binomials

Р 3 (x) = а(х – α)(х – β)(х – γ), or decomposes into the product of a binomial and a square trinomial Р 3 (x) = а(х – α)(х 2 + βх + γ ).

7. Any polynomial of the fourth degree can be expanded into the product of two square trinomials.

8. A polynomial f(x) is divisible by a polynomial g(x) without a remainder if there is a polynomial q(x) such that f(x) = g(x) · q(x). To divide polynomials, the “corner division” rule is used.

9. For the polynomial P(x) to be divisible by a binomial (x – c), it is necessary and sufficient that the number c be the root of P(x) (Corollary of Bezout’s theorem).

10. Vieta's theorem: If x 1, x 2, ..., x n are real roots of the polynomial

P(x) = a 0 x n + a 1 x n - 1 + ... + a n, then the following equalities hold:

x 1 + x 2 + … + x n = -a 1 /a 0,

x 1 x 2 + x 1 x 3 + ... + x n – 1 x n = a 2 /a 0,

x 1 x 2 x 3 + … + x n – 2 x n – 1 x n = -a 3 / a 0,

x 1 · x 2 · x 3 · x n = (-1) n a n / a 0 .

Solving Examples

Example 1.

Find the remainder of division P(x) = x 3 + 2/3 x 2 – 1/9 by (x – 1/3).

Solution.

Answer: R = 0.

Example 2.

Divide with a “corner” 2x 3 + 3x 2 – 2x + 3 by (x + 2). Find the remainder and incomplete quotient.

Solution:

2x 3 + 3x 2 – 2x + 3| x + 2

2x 3 + 4 x 2 2x 2 – x

X 2 – 2 x

Answer: R = 3; quotient: 2x 2 – x.

Basic methods for solving higher degree equations

1. Introduction of a new variable

The method of introducing a new variable is already familiar from the example of biquadratic equations. It consists in the fact that to solve the equation f(x) = 0, a new variable (substitution) t = x n or t = g(x) is introduced and f(x) is expressed through t, obtaining a new equation r(t). Then solving the equation r(t), the roots are found:

(t 1, t 2, …, t n). After this, a set of n equations q(x) = t 1 , q(x) = t 2 , … , q(x) = t n is obtained, from which the roots of the original equation are found.

Example 1.

(x 2 + x + 1) 2 – 3x 2 – 3x – 1 = 0.

Solution:

(x 2 + x + 1) 2 – 3(x 2 + x) – 1 = 0.

(x 2 + x + 1) 2 – 3(x 2 + x + 1) + 3 – 1 = 0.

Substitution (x 2 + x + 1) = t.

t 2 – 3t + 2 = 0.

t 1 = 2, t 2 = 1. Reverse substitution:

x 2 + x + 1 = 2 or x 2 + x + 1 = 1;

x 2 + x - 1 = 0 or x 2 + x = 0;

Answer: From the first equation: x 1, 2 = (-1 ± √5)/2, from the second: 0 and -1.

2. Factorization by grouping and abbreviated multiplication formulas

Example 1.

x 4 – 3x 2 + 4x – 3 = 0.

Solution.

Let's imagine - 3x 2 = -2x 2 – x 2 and group:

(x 4 – 2x 2) – (x 2 – 4x + 3) = 0.

(x 4 – 2x 2 +1 – 1) – (x 2 – 4x + 3 + 1 – 1) = 0.

(x 2 – 1) 2 – 1 – (x – 2) 2 + 1 = 0.

(x 2 – 1) 2 – (x – 2) 2 = 0.

(x 2 – 1 – x + 2)(x 2 – 1 + x - 2) = 0.

(x 2 – x + 1)(x 2 + x – 3) = 0.

x 2 – x + 1 = 0 or x 2 + x – 3 = 0.

Answer: There are no roots in the first equation, from the second: x 1, 2 = (-1 ± √13)/2.

3. Factorization by the method of undetermined coefficients

Example 1.

x 3 + 4x 2 + 5x + 2 = 0.

Solution.

A polynomial of degree 3 can be expanded into the product of linear and quadratic factors.

x 3 + 4x 2 + 5x + 2 = (x – a)(x 2 + bx + c),

x 3 + 4x 2 + 5x + 2 = x 3 +bx 2 + cx – ax 2 – abx – ac,

x 3 + 4x 2 + 5x + 2 = x 3 + (b – a)x 2 + (cx – ab)x – ac.

Having solved the system:

(b – a = 4,
(c – ab = 5,
(-ac = 2,

(a = -1,
(b = 3,
(c = 2, i.e.

x 3 + 4x 2 + 5x + 2 = (x + 1)(x 2 + 3x + 2).

The roots of the equation (x + 1)(x 2 + 3x + 2) = 0 are easy to find.

Answer: -1; -2.

4. Method of selecting a root using the highest and free coefficient

The method is based on the application of theorems:

1) Every integer root of a polynomial with integer coefficients is a divisor of the free term.

Example 1.

6x 3 + 7x 2 – 9x + 2 = 0.

Solution:

6: q = 1, 2, 3, 6.

Therefore, p/q = ±1, ±2, ±1/2, ±1/3, ±2/3, ±1/6.

Having found one root, for example – 2, we will find other roots using corner division, the method of indefinite coefficients or Horner’s scheme.

Answer: -2; 1/2; 1/3.

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Examples of solving equations with large degrees. Equations of higher degrees